   Chapter 11.8, Problem 22E

Chapter
Section
Textbook Problem

# Find the radius of convergence and interval of convergence of the series.22. ∑ n = 2 ∞ b n ln n ( x − a ) n ,   b > 0

To determine

To find: The radius of convergence and interval of convergence of the series is

Explanation

Given:

The series is n=2bnlnn(xa)n,b>0 ,

Result used:

(1)Ratio test: If limn|an+1an|=L<1 , then the series n=1an is absolutely convergent.

(2) The p-series n=11n is converges if p>1 and diverges if p1 .

(3)The comparison test:

“Suppose that an and bn are series with positive terms

(i)if bn is convergent and anbn for all n, then an is also convergent.

(ii) if bn is divergent and anbn for all n, then an is also divergent.”

(4) Alternating Series Test:

“If the alternating series n=1(1)n1bn=b1b2+b3b4+...   bn>0 satisfies, (i) bn+1bn   for all n and (ii) limnbn=0 , then the series is convergent”

Calculation:

Let an=bnlnn(xa)n

Then, an+1=bn+1ln(n+1)(xa)n+1 .

|an+1an|=|bn+1(xa)n+1ln(n+1)bn(xa)nlnn|

Taking limn on both sides

limn|an+1an|=limn|bn+1(xa)n+1ln(n+1)lnnbn(xa)n|=limn|bn+1(xa)n+1bn(xa)nlnnln(n+1)|=limnb|xa|(lnnln(n+1))

On simplify the logarithm function,

limnb|xa|.(lnnln(n+1))=b|xa|limn(lnnln(n+1))=b|xa|limn(1n1n+1)      (by L'Hospital's rule)=b|xa|limn(n+1n)=b|xa|limn(1+1n)

Apply the limit and simplify the terms as shown below

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